Question

infinite series

Answer 1

\[\sum_{n=0}^{oo}c^{n+1}=7\]

solve for c

Answer 2

this is a geometric series, no?

Answer 3

ya i think so

Answer 4

do you remember any formulas associated with geometric series? the infinite sum one would be handy ;)

Answer 5

i kinda do but it becomes a quadratic and doesnt make sense

Answer 6

well, let's work through it. Do you know the formula?

Answer 7

r=a_n/a_n-1

a/1-r

Answer 8

ok. so, we are saying that
7=a/(1-r) correct? from your original summation it seems as you don't actually have a constant factor a, so we can just take a=1
so then we have 7=1/(1-r)

Answer 9

can you solve for r?

Answer 10

ya

Answer 11

r=6/7

Answer 12

but thats not the answer i think because its not in the form c^(n-1)

Answer 13

lol one sec my brain is being silly

Answer 14

ah I see my indexing is off by one.

Answer 15

C<1

Answer 16

let's reindex our summation so it begins at zero.
\(\Large\sum_{n=0}^{\infty}c^{n}=7+c^0\)

Answer 17

wouldnt you divide/multiply the 7?

Answer 18

nvm dunno

Answer 19

\(\Large\sum_{n=0}^{\infty}c^{n+1}=7\)
\(\Large\sum_{n=0}^{\infty}(c^{n+1})+c^0=7+c^0\)
\(\Large\sum_{n=0}^{\infty}c^{n}=7+c^0\)

Answer 20

so then we have c^n=8

Answer 21

yup. so a=1, implying that
1/(1-r)=8 correct? let's solve for r then

Answer 22

r=7/8

Answer 23

that's what I got.
https://www.wolframalpha.com/input/?i=summation&a=*C.summation-_*Calculator.dflt-&f2=(7%2F8)%5En&f=Sum.sumfunction_(7%2F8)%5En&f3=1&f=Sum.sumlowerlimit_1&f4=infinity&f=Sum.sumupperlimit2%5Cu005finfinity&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*--

Answer 24

correct

Answer 25

thank you!

Answer 26

haha np - sorry for the slip up it's a bit late here :)

Answer 27

i suck really bad at these types of problems

Answer 28

yeah they confuse me too (I'm sure you can tell :P)